Question 5.33 from "Topology and its Applications" by Baesner is to compute the fundamental group of the torus ($T^2$) with $n$ points removed. I can "see" in my mind that if we remove one point we get a bouquet of two circles. Less clear is what happens when we remove two (or more) points. Any hints?
Algebraic Topology – Fundamental Group of a Torus with Points Removed
algebraic-topologyfundamental-groups
Related Solutions
We have two representations of the Klein bottle as a fundamental polygon. The first is:
And the second is formed by cutting this polygon across the diagonal, flipping one piece and reattaching it to give essentially two real projective planes glued together:
You should be able to see that as CW complexes and a 2-cell attached according to the diagram, these both form the Klein bottle with non-orientable genus 2.
Removing a point from this 2-cell produces a space that deformation rectacts onto the 1-skeleton, which in both cases obviously forms the wedge sum of two circles and the fundamental group is $\Bbb{Z} * \Bbb{Z}$.
Let's see if we can develop some sort of physical intuition for this. If the point (or by deformation, hole) we remove is in the right place, we can embed this in $\Bbb{R}^3$ to get a physical intuition.
Which then forms 
And you can see rather easily that this deforms to:
Which obviously has the fundamental group of $\Bbb{Z} * \Bbb{Z}$, as this deformation retracts onto $S^1 \vee S^1$.
The background to this question is discussed in this mathoverflow discussion. Since two points, say $a,b$, are to be identified, it is as well to start with a model of the whole situation, namely the fundamental groupoid $G=\pi_1(X,C)$ on the set $C$ consisting of the two points $a,b$. Since $X$, presumably the $2$-sphere, is simply connected, $G$ is isomorphic to the groupoid $\mathcal I$ which has two objects $0,1$ and exactly one arrow $\iota:0 \to 1$, as well as $\iota^{-1}: 1 \to 0$, and identities at $0,1$. When you identify $0,1$ of $\mathcal I$ to a single point in the category of groupoids you get the group $\mathbb Z$ of integers.
All this is explained in the book Topology and Groupoids, the third edition of a book published in 1968, i.e. nearly 50 years ago. A translation of some 1984 comments of Grothendieck on the neglect of this use of many base points is given here. See also the account by Philip Higgins in this now downloadable Categories and Groupoids (originallly 1971).
My understanding is that algebraic topology deals with algebraic modelling of geometric situations, and groupoids, which may have many objects, allow the modelling of the identification of several base points. If you are tethered to only one base point, then things are not so clear, and students may get confused.
May 27: Just to add some information. The key point is that is that the category $Gpd$ has a forgetful functor $Ob: Gpd \to Sets$ which is a "cofibration" of categories, meaning that if $G$ is a groupoid and $f: Ob(G) \to Y$ is a function then there is a groupoid $f_*(G)$ and morphism $U_f: G \to f_*(G)$ with a universal property that can be expressed in a pushout diagram of groupoids $$\begin{matrix} Ob(G) & \to & Y\\ \downarrow & & \downarrow\\ G & \to & f_*(G) \end{matrix} $$ The construction of $f_*(G)$ was introduced by Philip Higgins and includes that of free groups and free products of groups.
The relevance of this to the generalised van Kampen Theorem is given in Section 9.1 of Topology and Groupoids.
Best Answer
It's a bouquet of $n+1$ circles, so the free group on $n+1$ generators. Think of a rectangle with $n-1$ horizontal lines across it. Roll it up into a tube, so you have a line segment with $n+1$ circles attached to it. Then identify the top and bottom circles. So you have a circle with $n$ circles attached to it which is homotopic to a bouquet of $n+1$ circles.